Stochastic processes as Banach or Hilbert-space-valued random variables

Question

I'm having a hard time rectifying the following two concepts in my head.

A real-valued stochastic process is typically first described as a function

$$ X:\Omega\times T\rightarrow\Bbb{R} $$ where $(\Omega,\mathcal{F},\Bbb{P})$ is a probability space and $T$ is a parameter set, typically taken as a subset of $\Bbb{R}^d$ (though, for generalized processes, $T$ is a set of test functions). The probability measure $\Bbb{P}$ induces a probability measure $\Bbb{P}_X$ on the cylinder sets of $\Bbb{R}^T$ through the pushfoward. However, it is usually stated that this probability measure is fairly useless because the sigma algebra on $\Bbb{R}^T$ is somehow not 'rich' enough to ask interesting questions (for instance, the set of continuous functions is not measurable).

In other contexts - mainly engineering and in the field of uncertainty quantification - I see stochastic processes introduced in the following way. Fix a Banach or Hilbert space $\mathcal{Y}$ of functions on $T$ (such as $L^2(T)$ with $T\subset\Bbb{R}^d$) and make it into a measurable space by giving it the Borel sigma algebra; a 'stochastic process' is then assumed to be a $\mathcal{Y}$-valued random variable, that is,

$$ X:\Omega\rightarrow \mathcal{Y} $$ Now, we have that the law of $X$ is a Borel probability measure, which has (perhaps) nicer properties than in the $\Bbb{R}^T$ picture.

How can I connect these two pictures? Under what conditions can a stochastic process be assumed without loss of generality to be a Banach-space valued random variable?

I've browsed through several books and find discussions of both pictures, but nothing that really discusses the connection. Am I looking in the wrong place, or just being thick?

Edit: to rephrase, what I'm asking is this: are there any nice, general conditions under which the realizations of a classical or generalized random process $X$ (which are functions of $t\in T$, where $T$ might be a subset of $\Bbb{R}^d$ or a space of test functions) are almost surely in the specified Banach space $\mathcal{Y}$? It's very easy to find sufficient condition cases where it is easy to prove such a thing, for instance if $X$ has a continuous modification and $T$ is compact, then certainly $X_t \in L^p(T)$. But I'm interested in more general processes and random fields that might have discontinuous realizations and so forth, so I don't really care about sample path continuity, only sample path integrability.

Answer 1

A stochastic process, at an informal level, is a random object which changes in time. Being slightly more formal, we may write this as a collection $\{X_t\}_{t\in T}$ where $T$ is some index set which we should think of as time. We can therefore think of this as three things:

1. A function $T\times\Omega\to\mathbb R:(t,\omega)\mapsto X_t(\omega)$;
2. A function $\Omega\to S:\omega\mapsto\{X_t(\omega)\}_{t\in T}$, where $S\subseteq\mathbb R^T$ is some space of trajectories;
3. A function $T\to L:t\mapsto X_t$, where $L\subseteq\mathbb R^\Omega$ is some space of random variables.

Since we are studying probability theory, we need to impose some kind of measurability requirements in each case, and this ends up making the three definitions not equivalent and sometimes even not compatible. For instance, in 2, if $S=\mathbb R^T$ and we take the map to be measurable with respect to the product $\sigma$-field, then as you pointed out $\{\omega:t\mapsto X_t(\omega)\text{ is continuous}\}$ is not measurable (assuming $T=[0,1]$ or something like that). On the other hand, in 3 we could take $L=L^2(\Omega)$ which is an abuse of notation, since $L^2(\Omega)$ is really a space of equivalence classes, not functions, and thus is not a subset of $\mathbb R^\Omega$.

In some cases, though, any definition will do. Take $T=[0,1]$ and $X$ as Brownian motion, for instance. I've seen Brownian motion defined as a measure on $\mathbb R^T$, or on $C(T)$, or as a collection of $L^2$-random variables which happens to be continuous. In fact, different constructions of Brownian motion require interpretations.

The solution to this perceived problem is not to worry about it too much. Keep our original informalism at the back of your mind. Choose a definition and stick with it, and keep your assumptions precise in your mind, but also bear in mind what a stochastic process should be and don't let yourself get lost in notation.

Answer 2

As a partial answer to my own question, here is a proposition that is fairly easy to prove.

Suppose that $T\subset \Bbb{R}^d$, and the process is second order so that $\Bbb{E}\vert X(t)\vert^2<\infty$ for all $t\in T$. Suppose WLOG that the mean of $X$ is zero. Then, if the covariance function $K(s,t) = \Bbb{E}X(s)X(t)$ is integrable on the diagonal, the realizations of $X$ are almost surely square integrable with respect to $t$, that is $X(t,\omega)\in L^2(T)$ a.e. $\omega\in\Omega$. This is a simple consequence of Fubini-Tonelli:

$$ \Bbb{E}\|X(\cdot,\omega)\|_{L^2(T)}^2 = \Bbb{E}\int_T \vert X(t,\omega) \vert^2dt = \int_T \Bbb{E}\vert X(t,\omega)\vert^2 dt = \int_T k(t,t) dt $$ So, if the last integral is finite, $\|X(\cdot,\omega)\|_{L^2(T)}$ is finite for almost all $\omega\in\Omega$.

What if $X$ is a generalized process, so that (say) $T = \mathcal{D}$, the space of smooth compactly supported functions? (I realize this is a much harder question, which should probably be asked separately).